Columns fail by buckling when their critical load is reached. Long columns can be analysed with the Euler column formula
F = n π2 E I / L2 (1)
where
F = allowable load (lb, N)
n = factor accounting for the end conditions
E = modulus of elastisity (lb/in2, Pa (N/m2))
L = length of column (in, m)
I = Moment of inertia (in4, m4)
Factor Counting for End Conditions
- column pivoted in both ends : n = 1
- both ends fixed : n = 4
- one end fixed, the other end rounded : n = 2
- one end fixed, one end free : n = 0.25
Example - A Column Fixed in both Ends
An column with length 5 m is fixed in both ends. The column is made of an Aluminium I-beam 7 x 4 1/2 x 5.80 with a Moment of Inertia iy = 5.78 in4. TheModulus of Elasticity of aluminum is 69 GPa (69 109 Pa) and the factor for a column fixed in both ends is 4.
The Moment of Inertia can be converted to metric units like
Iy = 5.78 in4 (0.0254 m/in)4
= 241 10-8 m4
The Euler buckling load can then be calculated as
F = 4 π2 (241 10-8 m4) (69 109 Pa) / (5 m)2
= 262594 N
= 263 kN